Recent content by KokoDaSilvaback

So, the impedance of the parallel portion is ##(R^{-1}+ (jwl)^{-1})^{-1}##, correct? That preserves the real and imaginary components, which can then be added in series with the other impedance values.

So I'm back to square one essentially? Since the current will split between the parallel branches, there is no way 6.5mA will go to the resistor.

Ok, after rerunning the number and keeping my real portion separate from my imaginary portion, I end up with an impedance of 1528.72Ω, which when used to divide the 10VRMS, gives me IRMS = 6.54mA. That value closely matches the value sought, but do I now need to calculate the voltage drop across...

So rather than take the magnitude of the parallel impedance, simply add 1/R and 1/XL, then add those values to the rest of the reactance values, leaving the impedance calculation for the final step?

For a parallel circuit I have: $$Z = \frac {1} {\sqrt{(1/R)^2 + (1/jwl)^2}},$$ but when I perform that calculation, the inductance is no longer imaginary, as I have to use the complex conjugate in order to properly calculate the impedance of the parallel branch. The value I got from this...

So in that case would I treat the reactance (XL and XC) as entirely imaginary and the resistance as real? Once those are found, then calculate the impedance of the circuit?

Homework Statement Referencing the attached picture: a) Show that the RMS current in the 1 kΩ resistor is 6.5mA. b) If the AC voltage source was replaced by a battery, what would the current in the resistor be? Homework Equations V = IR Z = √(R^2 + X^2) The Attempt at a Solution To begin, I...